8. (a) IMPULSE = CHANGE IN MOMENTUM. See Example 8.2
and also note this problem is related to the last lab-- A VERNIER
LAB WILL HAVE A MINI LAB ON THIS SUBJECT.
Note:m*vf - m*vi = change in momentum
where vi and vf represent velocity
components along the x - axis. Let rightward be positive. For example
the pitched ball initially moves to the right, so vi
is positive and equals +45.0 m/s. After impact with the bat,
the ball moves left in the negative direction; thus vf
< 0 and equals - 55.0 m/s .
(b) Clearly the x-component of force Fx is
negative. We have m*vf - m*vi = Fx
*change in time. Note Fx is the average force.
Normally, Impulse =
m*vf - m*vi
= area under the force curve, obtained via integration.
But it can be shown that the area under the curve = average
force *(change in time) from the MEAN
VALUE THEOREM FOR INTEGRALS (Review Math 1) .

13.

(a) IMPULSE = CHANGE IN MOMENTUM. The problem seems to have
been plucked from EXP 20 VERNIER where you had to analyze
the shape of the force curve. In the experiment, clearly the shape
was not a perfect rectangle like this idealization , but the main
concepts apply: The area under the curve represents the
change in momentum resulting from the force. | m*vf -
m*vi | = area under the force curve. Note
we have given you the magnitude of the change in momentum or
impulse since the force curve plots the force magnitude.

(b) Assume the rightward direction is positive. You are given vi
= +5.00 m/s. To get the exact signed component of impulse
you would write:
m*vf - m*vi
= area under the force curve.

(i) If the force acts rightward m*vf
- m*vi = area under the force curve =
positive area of rectangle, in which case the final velocity
component will be greater than the initial velocity
component.
(ii) If the force acts leftward m*vf
- m*vi = area under the force curve =
negative area of rectangle, in which case the final velocity
component will be less than the initial velocity component ;
indeed vf could be negative depending on the
magnitude of the negative area but that is up to you to investigate

22. m1*v1i + m2*v2i
= m1*v1f + m2*v2f
. Let right be positive; thus v1i = + 1.50
m/s and v2i = -1.10 m/s. Also you are given that
v1f = + 0.250 m/s.
(a) Find velocity component of the second (2)
mass. Hint After collision the second mass is moving to the
right.
(b) There are two ways to explore this. One way is to find out
if the collision is elastic or not and possibly
bypass the kinetic energy computation. See equation
8.27 and discussions thereof and simply find out
if the magnitudes of the relative velocities are equal
before and after the collision. In other words , check
if : v1i - v2i = v2f
- v1f .
Note we used this concept in EXP 19 Vernier for the magnetic
bumper to magnetic bumper collisions. If the equality is
true, then the collision is elastic and the change in the
combined kinetic energy will be zero. If the equality is false,
then you must compute the change in combined kinetic energy in
the following way:
Change in combined kinetic energy = (1/2)*m1*v1f2 + (1/2)*m2*v2f2
- (1/2)* m1*v1i2 - (1/2)*m2*v2i2 .

24. (a) This part is essentially a replica of example 8.4. mA*vAi
+ mB*vBi = mA*vAf
+ mB*vBf , where
mA*vAi + mB*vBi
= 0, since the closed system starts from rest.
Thus, mA*vAf + mB*vBf
= 0 and you can now find the final velocity of A given
B's final velocity.

(b) Spring potential energy = (1/2)*k*x'2 = combined
kinetic energy after the the spring has expanded =
(1/2)*mA*vAf2 + (1/2)*mB*vBf2 . From this you can find the spring
potential energy. Note if I gave you the spring's force
constant k, you could find the distance |x'| it was
compressed.

28. This is a subtle variation of #24--- Here the system (
YOU AND THE ROCK) is also at rest initially but the rock has a
final velocity that is not collinear with the final velocity of
You as you slide in a direction opposite the horizontal component
of rock's final velocity.
m1*v1i + m2*v2i
= m1*v1f + m2*v2f
. which translates into:
0 = m1*v1f + m2*v2f
, where the written velocities are are x-components. Let 1 =
You and 2 = rock,
Clearly v2f = 12*cos 35. Find v1f
. We are assuming rightward is positive, so you will recoil
to the left in the negative direction.

Note we have assumed for this system, rock and You,
the net external force in the x direction is zero,
resulting in conservation of momentum in the x-direction. In other
words, in the x direction, the only relevatn forces acting
on system members You and rock are action reaction
pairs , which sum to zero.

In the y direction , momentum not conserved for the
system of You and rock since the rock has an upward y directed
velocity of 12*sin 35 and You have NO y-component of
velocity. This will be discussed in greater detail later, but the
bottom line is this: The rock transfers its y-component of
momentum to the Earth. So momentum is conserved in the y direction
if you expand the system to include the
Earth. In other words, it can
be easily be shown m2*12*sin 35 = momentum
transferred to Earth. Here is how:

ISOLATE THE ROCK: m2*a = pos - neg = FRH - m2g,
where FRH is the magnitude of the force on the
rock from the hand. We know m2*a = change
in momentum in y-direction/time , where time is the time period
your hand exerts a force on the rock. But we also know
change in momentum in y direction = m2*12*sin 35.
THUS:
m2*a = pos - neg = FRH - m2g or
FRH = m2g + m2*a
= m2g + m2*12*sin 35/time. Note
FRH is the normal force of the hand ( YOUR
HAND) on the rock.

ISOLATE YOU ( AND YOUR HAND) UNDER THE ROCK: Let's look at the situation before throw the rock.
Since your vertical acceleration a is zero we have:
sum of the force in y direction on you = pos - neg = 0 = N -
m1*g - m2*g since both the rock's
weight and your own weight act on you vertically down. Note
N is the magnitude of the normal force on you from the Earth.
Thus, N =
m1g + m2*g before you throw the
rock.
While you throw the rock, the following forces are acting on you:
Upward, the normal force of magnitude N, and downward, your
weight m1g + FHR. Note: FHR
= FRH = m2*g + m2*12*sin
35/time
Since N - m1*g - FHR
= 0, N = m1g
+ FHR = m1*g + m2*g
+ m2*12*sin 35/time. The the normal force has increased
by the term m2*12*sin 35/time. Finally throughout
all this discussion, the above normal force of magnitude N is
equal and opposite to the force on Earth of magnitude N' due to
you in contact with it. Thus N' = N quoted above.

We have the following data to compare with regard to the
normal force of magnitude N' on Earth exerted by you:

Before you throw rock: N' = m1*g + m2*g
= m1*g + FHR since FHR
= m2*g before throwing.
While you throw rock: N' = m1*g + FHR
= m1*g + m2*g + m2*12*sin
35/time. Now if you consider these forces acting down on the Earth
for the time period referred to as "time", the
time the rock takes to gain its momentum
m2*12*sin 35, then we would rewrite the
above equation for the total momentum transferred to
Earth during this interval:
TOTAL MOMENTUM = FORCE *TIME = N'*time = m1*g*time
+ m2*g*time + m2*12*sin 35.
The last term is the magnitude of downward momentum
transferred to Earth by rock you throw up with
vertical momentum
m2*12*sin 35. QED.

THE proof could repeated in the case of
"tryout" # 87, discussed below. But in future problems a
proof will normally not be required.

Solve these two equations simultaneously for VA
and VB. Then answer the other questions.

36. CLASS NOTES.

41. See example 8.9. After they collide, the momentum is (m + M)*Vf*
at an angle of 24 degrees with vertical or 66 degrees with the
horizontal . We have (m + M)*Vf cos 66 = m*Vc
and (m + M)*Vf sin 66 = M*VT where m is
the mass of the compact, M is the mass of the Truck and Vf
= 16.0 m/s. Find Vc andVT.